PHP - float numbers wrong precision with json_decode

Question

I have an issue when trying to decode JSON with the json_decode function in PHP 7.0.21. I am able to replicate this problem with these lines of code:

Code:

<?php
$inputJSON = '{"value":0.00000883}';
$outputJSON = json_decode($inputJSON);
print_r($outputJSON);

Output:

stdClass Object
(
    [value] => 8.83E-6
)

I also tried changing the precision with ini_set('precision', 8); which doesn't change the output.

The only fixes I found online were regex replaces which changed the number into a string but that's a hack and a good solution. I don't want to change ALL my float numbers to strings.

Why is this happening and how can I fix this properly without adding a lot of overhead like using number_format. Is the parsing simply broken in json_decode?

That's the same number. 8.83E-6 is 8.83 * 10(^-6), which means go with the decimal point 6 times to the left. — FirstOne
– FirstOne, Commented Nov 27, 2017 at 11:29
What actual problem is this causing? Those formats are functionally identical in PHP. — iainn
– iainn, Commented Nov 27, 2017 at 11:29
That's called scientific notation. There are a bunch of dups out here, and here is one of them: Why is PHP printing my number in scientific notation, when I specified it as .000021? — FirstOne
– FirstOne, Commented Nov 27, 2017 at 11:30
That's just the result of casting a PHP float variable into string to be printed on screen (that's what print_r() accomplishes). If it's a problem in your DBMS, you must be using prepared statements incorrectly (or not using them at all). We need to see the relevant code. — Álvaro González
– Álvaro González, Commented Nov 27, 2017 at 11:37
If your column requires that format then number_format is hardly a overhead. — Lawrence Cherone
– Lawrence Cherone, Commented Nov 27, 2017 at 11:38

IsThisJavascript · Accepted Answer · 2017-11-27 11:39:04Z

3

You can use number_format to remove the e-6 so you can store correctly in your database;

<?php

$inputJSON = '{"value":0.00000883}';
$outputJSON = json_decode($inputJSON);

$formatted = number_format($outputJSON->value,8);

print_r($formatted);

outputs: 0.00000883

Although, I'm pretty sure MySQL should handle 8.83E-6 as an input.

answered Nov 27, 2017 at 11:39

IsThisJavascript

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4 Comments

JohnMoe Over a year ago

This is indeed what fixes the display error on my part, I tried to look for an explanation at the wrong place while it was a database issue. Setting the ini_set('serialize_precision', -1); somehow fixed my database insertion problem.

Álvaro González Over a year ago

In general terms this should just not be necessary. Not only do the major database engines I've tested (MySQL, SQL Server and Oracle) accept 8.83E-6 as number literal but whatever PHP database library you are using should handle numeric parameters just fine.

alex Over a year ago

What would be the solutions if $inputJSON has multiple values to be formatted and I am returning an array $outputJSON = json_decode($inputJSON, true);

IsThisJavascript Over a year ago

Hi @alex, you'll only want to run number_format on integers/longs/doubles and floats. So it would be $formatted = number_format($outputJSON['TheArrayKey'],8); where TheArrayKey is your array key for one of the above variable types. Since you have multiple to format, I would either make an object class for this or just loop through the results and reset the array. Check out how to pass by reference

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